Why does a freewheeling flywheel counter-rotate slightly before finally stopping?

My question is this: why does a spinning flywheel, supported by reasonably sensible bearings* that is allowed to coast down to rest, rock backwards slightly before finally stopping completely? I assume that it doesn't come to rest after the first "rock" or reversal of direction, rather it decays.

*I mention this because I haven't observed it when the bearings of whatever it is are particularly tight or knackered, or there is a lot of parasitic drag, such as when the flywheel has an engine attached. As you'd probably expect.

I have a rubbish video: https://www.youtube.com/watch?v=khXpTaNs9Fw

That's a Technics 1210 with the magnet removed. I removed it for the video to make sure that the rocking at the end of the coasting wasn't caused by the back EMF induced in the motor.

I tried to illustrate that it does it in both directions.

I've been wondering about the zero crossing point, or the point in time between (arbitrarily) clockwise rotation and anticlockwise rotation. In my mind, if the reason it does this is simply that the friction is of a higher magnitude than the inertia (it has to be, otherwise it would not decelerate if it were of an equal magnitude and would accelerate if it were of a lower magnitude) then how can it be of a magnitude significant to overcome the reverse static friction, which is higher than the dynamic friction seen during the deceleration?

• I don't know much about flywheels. Are you sure it isn't just because the wheel is slightly unbalanced? – Brian Moths Jan 9 '17 at 22:29
• I checked for that. It does it no matter where I start the wheel and no matter where it stops, no matter which direction. I also tilted the turntable and it still does it. – mc172 Jan 9 '17 at 22:32
• I've deleted your video link because it caused problems on my laptop and I'm unlikely to be the only one to experience that. Please do not reinstate it. Post as a YouTube or another safe format, please. Thank you. – Gert Jan 9 '17 at 22:37
• "Causing problems" doesn't really give me much of a chance of finding a suitable alternative. What problems exactly? – mc172 Jan 9 '17 at 22:47
• About the video link, I don't think it really matters what the problem was. My experience was I opened the page in a new tab, looked at the website for about three seconds, saw that it wanted me to download something, then closed the tab. I think people would be more comfortable with youtube or another familiar streaming site. – Brian Moths Jan 9 '17 at 22:57

When a flywheel decelerates from its initial angular velocity $\omega_0$ to standstill, that is caused by torque $\tau$, acc. Newton:

$$\tau=I\alpha,$$

where $I$ is the inertia moment of the flywheel and $\alpha$ the angular deceleration:

$$\alpha=-\frac{\mathrm{d}\omega}{\mathrm{d}t}$$

(The minus sign represents deceleration)

If for simplicity's sake we assume $\tau$ to be constant, then:

$$\alpha=-\frac{I}{\tau}$$

And the angular velocity $\omega(t)$, as a function of time:

$$\omega(t)=\omega_0-\frac{I}{\tau}\Delta t$$

If torque persists then $\omega \leq 0$ when:

$$\Delta t \geq \frac{\omega_0 \tau}{I}$$

So this is the moment in time the flywheel's sense of rotation would reverse.

Now, if we consider a freewheeling flywheel with no external, intentional braking torque imposed on it, then the only phenomenon that can reduce the angular velocity is friction. Even the best constructed flywheel experiences some friction in the bearings, as well as some air drag. These forces do provide a torque, in the opposite sense the of the motion and this works to reduce the angular velocity, as described above.

However, friction forces always act in the opposite sense of motion and decay completely when motion decays. Mathematically:

$$\omega=0 \implies \tau_{\mathrm{friction}}=0$$

This means that friction force (torque, to be precise) can never be responsible for any reversal of the sense of rotation.

But if we look at the case of the turntable (YouTube in the OP's comment), the required reversing torque is quite easily explained.

The 'secret' lies in the turntable's drive belt. The drive belt is not perfectly unstretchable (ineleastic, usually they're made of rubber, sometimes fabric reinforced).

When the turntable comes to a halt, the drive belt becomes slightly stretched, making it act like a stretched spring. Like a spring it now provides a small restoring torque, acting in the opposite sense of the original sense of rotation. As above, this torque now causes angular acceleration in the opposite sense of the original sense of rotation, causing the brief reversal.

A very related way of looking at it is that the drive belt stores a small amount of potential energy $U$. during the final stages of braking. This is then converted to rotational kinetic energy $K$:

$$U=K=\frac12 I\omega_1^2,$$

where $\omega_1$ is the angular velocity at the end of the reversal.

'On paper' the turntable should enter an oscillation (you can kind of see it in the video) but friction causes that motion to cease quickly because friction expends work (energy).

• +1. I think the calculation only gets in the way of the point you make in the 1st part of your answer, and detracts from the 2nd part, which convinces on its own. – sammy gerbil Jan 10 '17 at 0:12
• Going by the OP's question I felt it was important to make the point that friction alone cannot explain the reversal (it's a commonly held misconception about friction). Thanks for the upvote, Sammy! – Gert Jan 10 '17 at 0:17
• Thanks for the detailed reply, Gert. The turntable has no drive belt - an omission on my own part. Please consider it as a flywheel on a fairly low-friction bearing. The motor is a direct drive one, and I have removed the magnetics, so that no back EMF can be generated to affect the result. – mc172 Jan 10 '17 at 0:23
• @mc172: this means that the elastic potential energy I wrote about is stored in the direct drive during braking. Where exactly is hard to tell w/o precise information on the drive's specifics. Most things are eleastic to various extents. – Gert Jan 10 '17 at 0:32
• Here is a video of the bearing: youtube.com/watch?v=iyzykI3sJlw That's all that was in there when I made the video in the original post. The platter is attached to that bearing by a taper. The axial thrust is taken by the hemispherical geometry on the end of the shaft (seen at around 0:45) and radial loads are taken by the bronze bush. Here's under the platter: youtu.be/hS1w0hA01go?t=78 The ring underneath the platter is the magnet I removed. You can see the stator under the platter briefly in this video. – mc172 Jan 10 '17 at 0:40

Here is another possible explanation of what you observed. For this explanation to work, two conditions are necessary: 1) Center of gravity of flywheel is off-center from its axis of rotation. 2) Plane of flywheel is not exactly horizontal. In a real flywheel it is very likely that both of these conditions are satisfied.

If this is the case then for small angular displacements, flywheel behaves like a pendulum whose energy is continuously dissipated away by friction until it comes to rest. Initially of course you give it enough kinetic energy that it rotates about its hinge point, but after enough of its kinetic energy has been dissipated away so that it cannot complete a rotation, it will oscillate like a pendulum.

• I can test this by tilting the turntable and observing. I didn't spend too long messing around the first time and only tilted it a few (10?) degrees so I'll have another go. The platter isn't perfectly balanced but if I tilt it up to perpendicular it still doesn't move on its own to the low point. You have to get a feel carefully for where the low point is. What I'm not sure about is that with the platter fairly horizontal, as in roughly level by eye from arms length away, I wouldn't expect the magnitude of the pendulum effect to be large enough to bump the platter around like in my video. – mc172 Jan 18 '17 at 23:43
• @mc172 It is possible that my hypothesis is incorrect. Only experiments can tell. – Deep Jan 19 '17 at 5:02

This elastic 'rebound' is possibly from a viscous bearing, or more likely from the peripheral air layers. (I know from smoke experiments you can see a vortex above a turntable.)

Providing the peripheral air drag is more significant than the bearing friction, the turntable will slow down mainly because the air is dragging on it. This spinning 'hoop' of air around the turntable edge must therefore come to a standstill (relative to the deck) before the turntable does. The turntable now overshoots a little, only now it is 'stretching' the air layer, and when it finally stops, the turntable experiences a slight reverse airflow.

I now expect to see streamlining kits being sold to DJs to give greater precision when cueing!!

• I'm not sure if this really makes sense. It's the relative difference in air speed that slows you down; but I don't see how it would ever get to the point where it could cause backwards rotation. – JMac Mar 7 '18 at 20:49
• a) Remember air does not work exactly like a liquid. b) If you slow down a flywheel with a rubber band on its periphery as a constant-force brake, the amount of 'stretch' in the rubber will remain constant until the flywheel actually stops, whereupon the rubber band will drag the flywheel backwards until the rubber is relaxed. c) We know the real case is not constant-force, but there is a close approximation to a straight-line at the slowest speeds. – OookLout Nov 8 '18 at 23:12

It can be result of turning of bearing mounting. During slowing down of rotation, a continous torgue in bearing make a small angle displacement of all unit = bearing and its mounting in the direction of turntable rotation. When the turntable stops, then mounting starts to return to its neutral position, which is a litle angle back. --Another similiar problem: I saw today prof Lewin video https://www.youtube.com/watch?v=20IerfdG4Fs&list=PLyQSN7X0ro23CEzKOjAVcRq66m6g-mFLe . Very interesting. I can't understand his problem. Maybe elastic material??